\(\int \frac {x^4}{\sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\) [1444]

Optimal result

Integrand size = 24, antiderivative size = 164 \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {2}{45} x \left (2-3 x^2\right )^{3/4}+\frac {4 \sqrt [4]{2} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt {3} x \sqrt [4]{2-3 x^2}}{2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}} \] Output:

2/45*x*(-3*x^2+2)^(3/4)+4/27*2^(1/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2 
)^(1/2))*3^(1/2)/x/(-3*x^2+2)^(1/4))*3^(1/2)+4/27*2^(1/4)*arctanh(3^(1/2)* 
x*(-3*x^2+2)^(1/4)/(2^(3/4)+2^(1/4)*(-3*x^2+2)^(1/2)))*3^(1/2)-16/45*2^(1/ 
4)*EllipticE(sin(1/2*arcsin(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 5.90 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {1}{45} x \left (3\ 2^{3/4} x^2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\frac {2 \left (2-3 x^2+\frac {32 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{\left (-4+3 x^2\right ) \left (4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )\right )}\right )}{\sqrt [4]{2-3 x^2}}\right ) \] Input:

Integrate[x^4/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]
 

Output:

(x*(3*2^(3/4)*x^2*AppellF1[3/2, 1/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4] + (2*(2 
 - 3*x^2 + (32*AppellF1[1/2, 1/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4])/((-4 + 3* 
x^2)*(4*AppellF1[1/2, 1/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4] + x^2*(2*AppellF1 
[3/2, 1/4, 2, 5/2, (3*x^2)/2, (3*x^2)/4] + AppellF1[3/2, 5/4, 1, 5/2, (3*x 
^2)/2, (3*x^2)/4])))))/(2 - 3*x^2)^(1/4)))/45
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {349, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx\)

\(\Big \downarrow \) 349

\(\displaystyle \int \left (-\frac {x^2}{3 \sqrt [4]{2-3 x^2}}-\frac {4}{9 \sqrt [4]{2-3 x^2}}+\frac {16}{9 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}}+\frac {4 \sqrt [4]{2} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {2}{45} \left (2-3 x^2\right )^{3/4} x\)

Input:

Int[x^4/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]
 

Output:

(2*x*(2 - 3*x^2)^(3/4))/45 + (4*2^(1/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 
 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*Sqrt[3]) + (4*2^(1/4)*ArcTanh[ 
(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*Sqr 
t[3]) - (16*2^(1/4)*EllipticE[ArcSin[Sqrt[3/2]*x]/2, 2])/(15*Sqrt[3])
 

Defintions of rubi rules used

rule 349

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol 
] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(1/4)*(c + d*x^2)), x], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In 
tegerQ[m/2])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {x^{4}}{\left (-3 x^{2}+2\right )^{\frac {1}{4}} \left (-3 x^{2}+4\right )}d x\]

Input:

int(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)
 

Output:

int(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)
 

Fricas [F]

\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="fricas")
 

Output:

integral((-3*x^2 + 2)^(3/4)*x^4/(9*x^4 - 18*x^2 + 8), x)
 

Sympy [F]

\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=- \int \frac {x^{4}}{3 x^{2} \sqrt [4]{2 - 3 x^{2}} - 4 \sqrt [4]{2 - 3 x^{2}}}\, dx \] Input:

integrate(x**4/(-3*x**2+2)**(1/4)/(-3*x**2+4),x)
 

Output:

-Integral(x**4/(3*x**2*(2 - 3*x**2)**(1/4) - 4*(2 - 3*x**2)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="maxima")
 

Output:

-integrate(x^4/((3*x^2 - 4)*(-3*x^2 + 2)^(1/4)), x)
 

Giac [F]

\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}} \,d x } \] Input:

integrate(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="giac")
 

Output:

integrate(-x^4/((3*x^2 - 4)*(-3*x^2 + 2)^(1/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\int \frac {x^4}{{\left (2-3\,x^2\right )}^{1/4}\,\left (3\,x^2-4\right )} \,d x \] Input:

int(-x^4/((2 - 3*x^2)^(1/4)*(3*x^2 - 4)),x)
 

Output:

-int(x^4/((2 - 3*x^2)^(1/4)*(3*x^2 - 4)), x)
 

Reduce [F]

\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\left (\int \frac {x^{4}}{3 \left (-3 x^{2}+2\right )^{\frac {1}{4}} x^{2}-4 \left (-3 x^{2}+2\right )^{\frac {1}{4}}}d x \right ) \] Input:

int(x^4/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)
 

Output:

 - int(x**4/(3*( - 3*x**2 + 2)**(1/4)*x**2 - 4*( - 3*x**2 + 2)**(1/4)),x)